I'm an italian student trying to study bash. I have a question. I've to solve this following bash script
#!/bin/bash
echo hello
if [ -x "$namefile" ] ; then
echo "file $namefile exists"
else
echo " file $namefile doesn't exists"
fi
Reading your posts I understand that -x means if the file exists and is executable is true.
Now I have created a file with all permissions for all with chmod 777 file and I try to launch the script but the result is "File doesn't exists". Why?
This is a reverse engineering exercise and I can't modify the script.
If you don't want to modify your script, set the variable when calling the script:
namefile=myfile.txt ./sestoscr
(assuming that sestoscr is the name of the script you posted, and that the script has execute permission, and that myfile.txt is the name of the file you want to test using your script).
A more flexible solution would be to change the test in your script to
if [ -x "${namefile:=$1}" ] ; then
This still allows you to invoke the script via a change in environment (as outline above), but you also can call it using an explicit parameter,
./sestoscr myfile.txt
However, in this case I would give the positional parameter priority over the environment (because this is what most users would expect), and write it as:
namefile=${1:-$namefile}
if [ -x "$namefile" ]; ...
With this approach, if a parameter is supplied, it uses the parameter, and if none is supplied, it uses the environment variable namefile.
I have shell script which collects user input and generate a json file from the input. The input file is then read by a compiled matlab program which then creates an output file and I use that output file to plot.
I'm working on a linux server where I installed matlab runtime. Everything works fine inside the server, once I try to access the site outside the server the .sh doesn't completely execute( it only echoes the first lines and stops).
#!/bin/sh
# script for execution of deployed applications
#
# Sets up the MATLAB Runtime environment for the current $ARCH and executes
# the specified command.
#
exe_name=$0
exe_dir=`dirname "$0"`
echo "------------------------------------------"
if [ "x$1" = "x" ]; then
echo Usage:
echo $0 \<deployedMCRroot\> args
else
echo Setting up environment variables
MCRROOT="$1"
echo ---
LD_LIBRARY_PATH=.:${MCRROOT}/runtime/glnxa64 ;
LD_LIBRARY_PATH=${LD_LIBRARY_PATH}:${MCRROOT}/bin/glnxa64 ;
LD_LIBRARY_PATH=${LD_LIBRARY_PATH}:${MCRROOT}/sys/os/glnxa64;
LD_LIBRARY_PATH=${LD_LIBRARY_PATH}:${MCRROOT}/sys/opengl/lib/glnxa64;
export LD_LIBRARY_PATH;
echo LD_LIBRARY_PATH is ${LD_LIBRARY_PATH};
shift 1
args=
while [ $# -gt 0 ]; do
token=$1
args="${args} \"${token}\""
shift
done
eval "\"${exe_dir}/main\"" $args
fi
exit
Using a script located within your web service document root. As an example, this is a simple PHP file to run the shell script. The chdir path used is just an example:
<?php
chdir('/var/www/yourdir');
$output = shell_exec('./matlab_setup.sh arg1');
Note that making the script accessible via the internet means anyone using a browser can access it, unless you introduce futher safeguarding.
So, I'm guessing this may be a bug, or maybe I've botched something up, I dunno...
I have a script that was always working but I've been tasked to make it work using the new WSL2. I've snipped out the first block of code as it's giving me issues right off the bat. As you'll see below, I'm simply trying to determine if a variable has been set or not. This works in my Linux VM and also in Cygwin, however, it doesn't work in WSL2. Here is the code:
#!/bin/bash
echo $test_var
set -e
if [ ! -d "$test_var" ]; then
echo Please set test_var.
exit
fi
When I run this in any of the working systems I get the output of the variable.
In WSL2 I get the output of the variable followed by Please set test_var. And it stops the script due to the set -e as it's supposed to do thinking the var isn't set.
Any help would be appreciated.
Thanks
If your intention is to check if a directory exists (whose name apparently needs to be set as an environment variable in $test_var), if that directory exists relative to the current directory where the script is executed, you want something like:
#!/bin/bash
echo "$test_var"
set -e
if [ ! -d "$test_var" ]; then
echo "cannot find directory $test_var"
exit
fi
Note that here I have only changed your message, and your problem might possibly be explained by the fact that you do not have such a directory under WSL2.
If, on the other hand, you want to check if some environment variable (whatever it represents) is set, you want the -z option, something like:
#!/bin/bash
echo "$test_var"
set -e
if [ -z "$test_var" ]; then
echo "Please set test_var."
exit
fi
Note the absence of your negation sign (!).
I'm trying to change bundle display name via Xcode run script if a special condition is met. So far I have this:
if [ MY_CONDITION ]; then
BUNDLE_DISPLAY_NAME = ${BUNDLE_VERSION};
fi
I get this error
line 3: BUNDLE_DISPLAY_NAME: command not found
Where do I look up this fancy variable names? And is what I'm trying to do even possible with script?
Remove spaces around = symbol (BUNDLE_DISPLAY_NAME=${BUNDLE_VERSION};) otherwise bash interpret BUNDLE_DISPLAY_NAME as separate command and try to execute BUNDLE_DISPLAY_NAME, but not found this command. But bash interpret BUNDLE_DISPLAY_NAME=${BUNDLE_VERSION} as operation assign a value ${BUNDLE_VERSION} with variable BUNDLE_DISPLAY_NAME.
For clarity and to officially put #trojanfoe's comment in an answer, BUNDLE_DISPLAY_NAME isn't a build variable. It's a value in your target's Info.plist. You'll need to change it there instead.
In order to keep your source control clean, you should put the following script after your Copy Bundle Resources phase.
if [ MY_CONDITION ]; then
newDisplayName="${BUNDLE_VERSION}"
command="Set :CFBundleDisplayName $newDisplayName"
echo "Updating display name in app package to \"$newDisplayName\""
/usr/libexec/PlistBuddy -c "$command" "${TARGET_BUILD_DIR}/${INFOPLIST_PATH}"
filepath="${BUILT_PRODUCTS_DIR}/${WRAPPER_NAME}.dSYM/Contents/Info.plist"
if [ -f "$filepath" ]; then
echo "Updating display name in dsym to \"$newDisplayName\""
/usr/libexec/PlistBuddy -c "$command" "$filepath"
fi;
fi
Recently I wrote a script which sets an environment variable, take a look:
#!/bin/bash
echo "Pass a path:"
read path
echo $path
defaultPath=/home/$(whoami)/Desktop
if [ -n "$path" ]; then
export my_var=$path
else
echo "Path is empty! Exporting default path ..."
export my_var=$defaultPath
fi
echo "Exported path: $my_var"
It works just great but the problem is that my_var is available just locally, I mean in console window where I ran the script.
How to write a script which allow me to export global environment variable which can be seen everywhere?
Just run your shell script preceded by "." (dot space).
This causes the script to run the instructions in the original shell. Thus the variables still exist after the script finish
Ex:
cat setmyvar.sh
export myvar=exists
. ./setmyvar.sh
echo $myvar
exists
Each and every shell has its own environment. There's no Universal environment that will magically appear in all console windows. An environment variable created in one shell cannot be accessed in another shell.
It's even more restrictive. If one shell spawns a subshell, that subshell has access to the parent's environment variables, but if that subshell creates an environment variable, it's not accessible in the parent shell.
If all of your shells need access to the same set of variables, you can create a startup file that will set them for you. This is done in BASH via the $HOME/.bash_profile file (or through $HOME/.profile if $HOME/.bash_profile doesn't exist) or through $HOME/.bashrc. Other shells have their own set of startup files. One is used for logins, and one is used for shells spawned without logins (and, as with bash, a third for non-interactive shells). See the manpage to learn exactly what startup scripts are used and what order they're executed).
You can try using shared memory, but I believe that only works while processes are running, so even if you figured out a way to set a piece of shared memory, it would go away as soon as that command is finished. (I've rarely used shared memory except for named pipes). Otherwise, there's really no way to set an environment variable in one shell and have another shell automatically pick it up. You can try using named pipes or writing that environment variable to a file for other shells to pick it up.
Imagine the problems that could happen if someone could change the environment of one shell without my knowledge.
Actually I found an way to achieve this (which in my case was to use a bash script to set a number of security credentials)
I just call bash from inside the script and the spawned shell now has the export values
export API_USERNAME=abc
export API_PASSWORD=bbbb
bash
now calling the file using ~/.app-x-setup.sh will give me an interactive shell with those environment values setup
The following were extracted from 2nd paragraph from David W.'s answer: "If one shell spawns a subshell, that subshell has access to the parent's environment variables, but if that subshell creates an environment variable, it's not accessible in the parent shell."
In case a user need to let parent shell access your new environment variables, just issue the following command in parent shell:
source <your_subshell_script>
or using shortcut
. <your_subshell_script>
You got to add the variable in your .profile located in /home/$USER/.profile
Yo can do that with this command:
echo 'TEST="hi"' >> $HOME/.profile
Or by edit the file with emacs, for example.
If you want to set this variable for all users, you got to edit /etc/profile (root)
There is no global environment, really, in UNIX.
Each process has an environment, originally inherited from the parent, but it is local to the process after the initial creation.
You can only modify your own, unless you go digging around in the process using a debugger.
write it to a temporary file, lets say ~/.myglobalvar and read it from anywhere
echo "$myglobal" > ~/.myglobalvar
Environment variables are always "local" to process execution the export command allow to set environment variables for sub processes. You can look at .bashrc to set environment variables at the start of a bash shell. What you are trying to do seems not possible as a process cannot modify (or access ?) to environment variables of another process.
You can update the ~/.bashrc or ~/.bash_profile file which is used to initialize the environment.
Take a look at the loading behavior of your shell (explained in the manpage, usually referring to .XXXshrc or .profile). Some configuration files are loaded at login time of an interactive shell, some are loaded each time you run a shell. Placing your variable in the latter might result in the behavior you want, e.g. always having the variable set using that distinct shell (for example bash).
If you need to dynamically set and reference environment variables in shell scripts, there is a work around. Judge for yourself whether is worth doing, but here it is.
The strategy involves having a 'set' script which dynamically writes a 'load' script, which has code to set and export an environment variable. The 'load' script is then executed periodically by other scripts which need to reference the variable. BTW, the same strategy could be done by writing and reading a file instead of a variable.
Here's a quick example...
Set_Load_PROCESSING_SIGNAL.sh
#!/bin/bash
PROCESSING_SIGNAL_SCRIPT=./Load_PROCESSING_SIGNAL.sh
echo "#!/bin/bash" > $PROCESSING_SIGNAL_SCRIPT
echo "export PROCESSING_SIGNAL=$1" >> $PROCESSING_SIGNAL_SCRIPT
chmod ug+rwx $PROCESSING_SIGNAL_SCRIPT
Load_PROCESSING_SIGNAL.sh (this gets dynamically created when the above is run)
#!/bin/bash
export PROCESSING_SIGNAL=1
You can test this with
Test_PROCESSING_SIGNAL.sh
#!/bin/bash
PROCESSING_SIGNAL_SCRIPT=./Load_PROCESSING_SIGNAL.sh
N=1
LIM=100
while [ $N -le $LIM ]
do
# DO WHATEVER LOOP PROCESSING IS NEEDED
echo "N = $N"
sleep 5
N=$(( $N + 1 ))
# CHECK PROCESSING_SIGNAL
source $PROCESSING_SIGNAL_SCRIPT
if [[ $PROCESSING_SIGNAL -eq 0 ]]; then
# Write log info indicating that the signal to stop processing was detected
# Write out all relevent info
# Send an alert email of this too
# Then exit
echo "Detected PROCESSING_SIGNAL for all stop. Exiting..."
exit 1
fi
done
~/.bin/SOURCED/lazy script to save and load data as flat files for system.
[ ! -d ~/.megadata ] && mkdir ~/.megadata
function save_data {
[ -z "$1" -o -z "$2" ] && echo 'save_data [:id:] [:data:]' && return
local overwrite=${3-false}
[ "$overwrite" = 'true' ] && echo "$2" > ~/.megadata/$1 && return
[ ! -f ~/.megadata/$1 ] && echo "$2" > ~/.megadata/$1 || echo ID TAKEN set third param to true to overwrite
}
save_data computer engine
cat ~/.megadata/computer
save_data computer engine
save_data computer megaengine true
function get_data {
[ -z "$1" -o -f $1 ] && echo 'get_data [:id:]' && return
[ -f ~/.megadata/$1 ] && cat ~/.megadata/$1 || echo ID NOT FOUND
:
}
get_data computer
get_data computer
Maybe a little off topic, but when you really need it to set it temporarily to execute some script and ended up here looking for answers:
If you need to run a script with certain environment variables that you don't need to keep after execution you could do something like this:
#!/usr/bin/env sh
export XDEBUG_SESSION=$(hostname);echo "running with xdebug: $XDEBUG_SESSION";$#
In my example I just use XDEBUG_SESSION with a hostname, but you can use multiple variables. Keep them separated with a semi-colon. Execution as follows (assuming you called the script debug.sh and placed it in the same directory as your php script):
$ debug.sh php yourscript.php